Problem 93 - Chemistry: The Central Science

Question

Cyclopentadiene \(\left(\mathrm{C}_{5} \mathrm{H}_{6}\right)\) reacts with itself to form dicyclopentadiene \(\left(\mathrm{C}_{10} \mathrm{H}_{12}\right)\). A \(0.0400 \mathrm{M}\) solution of \(\mathrm{C}_{5} \mathrm{H}_{6}\) was monitored as a function of time as the reaction \(2 \mathrm{C}_{5} \mathrm{H}_{6} \longrightarrow \mathrm{C}_{10} \mathrm{H}_{12}\) proceeded. The following data were collected: \begin{tabular}{rl} \hline Time (s) & {\(\left[\mathbf{C}_{5} \mathrm{H}_{6}\right](M)\)} \\ \hline \(0.0\) & \(0.0400\) \\ \(50.0\) & \(0.0300\) \\ \(100.0\) & \(0.0240\) \\ \(150.0\) & \(0.0200\) \\ \(200.0\) & \(0.0174\) \\ \hline \end{tabular} Plot \(\left[\mathrm{C}_{5} \mathrm{H}_{6}\right]\) versus time, \(\ln \left[\mathrm{C}_{5} \mathrm{H}_{6}\right]\) versus time, and \(1 /\left[\mathrm{C}_{5} \mathrm{H}_{6}\right]\) versus time. What is the order of the reaction? What is the value of the rate constant?

Step-by-Step Solution

Answer

The order of the reaction can be determined by analyzing the plots. In this case, the straight line plot is the plot of \( \ln[\mathrm{C}_{5} \mathrm{H}_{6}] \) vs. time, which indicates the reaction is first-order. To determine the rate constant, k, divide the negative slope of the straight line plot by 2.

Step 1: Creating the plots

To determine the order of the reaction, create three separate plots: \( [\mathrm{C}_{5} \mathrm{H}_{6}] \) vs. time, \( \ln[\mathrm{C}_{5} \mathrm{H}_{6}] \) vs. time, and \( \frac{1}{[\mathrm{C}_{5} \mathrm{H}_{6}]} \) vs. time. It is suggested to use graphing software, a spreadsheet program, or a graphing calculator to effectively create the plots using the given data.

Step 2: Analyzing the plots

Once the plots have been created, analyze them to determine which plot shows a straight line. The straight line plot corresponds to the order of the reaction: - If the plot of \( [\mathrm{C}_{5} \mathrm{H}_{6}] \) vs. time is a straight line, the reaction is zero-order. - If the plot of \( \ln[\mathrm{C}_{5} \mathrm{H}_{6}] \) vs. time is a straight line, the reaction is first-order. - If the plot of \( \frac{1}{[\mathrm{C}_{5} \mathrm{H}_{6}]} \) vs. time is a straight line, the reaction is second-order. By examining the resulting plots, the order of the reaction can be determined.

Step 3: Determining the rate constant

Once the order of the reaction has been determined, calculate the rate constant, k, using the slope of the straight line plot. - If the reaction is zero-order, the rate constant can be found by dividing the negative slope of the plot by 2, as the reaction is \(2\mathrm{C}_{5}\mathrm{H}_{6} \longrightarrow\mathrm{C}_{10}\mathrm{H}_{12}\). - If the reaction is first-order, the rate constant can be found by dividing the negative slope of the plot by 2, following the same logic. - If the reaction is second-order, the rate constant can be found directly from the positive slope of the plot, without dividing by 2. By calculating the rate constant using the slope along with the order of the reaction, the student has successfully completed the exercise.

Key Concepts

Reaction Order

In the study of chemical reactions, understanding the reaction order is crucial as it defines how the concentration of reactants affects the rate of the reaction. To determine the reaction order, experimental data regarding the concentration of reactants over time is analyzed. Typically, data is plotted in three different forms:

• Concentration vs. time (\([ ext{C}_5 ext{H}_6]\) vs. time)
• Natural logarithm of concentration vs. time (\( ext{ln}[ ext{C}_5 ext{H}_6]\) vs. time)
• Inverse concentration vs. time (\(1/[ ext{C}_5 ext{H}_6]\) vs. time)

The plot that yields a straight line indicates the order of the reaction. For example:

• A straight line in the \([ ext{C}_5 ext{H}_6]\) vs. time plot suggests a zero-order reaction.
• A straight line in the \( ext{ln}[ ext{C}_5 ext{H}_6]\) vs. time plot suggests a first-order reaction.
• A straight line in the \(1/[ ext{C}_5 ext{H}_6]\) vs. time plot suggests a second-order reaction.

By identifying the reaction order, chemists can better understand the underlying mechanism and dynamics of the reactions they are studying.
Being systematic in plotting these data forms is essential for correct interpretation of reaction orders.

Rate Constant

Once the reaction order is determined, the next step is to calculate the rate constant, often denoted by the letter \(k\). The rate constant is a proportionality factor in the equation of the reaction rate and provides insight into the speed of the reaction. Calculating \(k\) depends on the reaction order that has been identified:

• For a zero-order reaction, \(k\) is equal to the negative slope of the \([ ext{C}_5 ext{H}_6]\) vs. time plot, divided by 2.
• For a first-order reaction, \(k\) is the negative slope of the \( ext{ln}[ ext{C}_5 ext{H}_6]\) vs. time plot, again divided by 2.
• For a second-order reaction, \(k\) can be directly taken from the positive slope of the \(1/[ ext{C}_5 ext{H}_6]\) vs. time plot, without division.

The division by 2 in zero and first-order reactions is due to the stoichiometry of the reaction where two molecules of cyclopentadiene react to form one dicyclopentadiene. The units of \(k\) will change depending on the order:

• Zero-order: \( ext{M/s}\)
• First-order: \(1/ ext{s}\)
• Second-order: \( ext{M}^{-1} ext{s}^{-1}\)

Understanding the rate constant not only helps to quantify the rate at which a reaction proceeds but also to compare the efficiency and speed of similar reactions under different conditions.

Integrated Rate Laws

Integrated rate laws provide mathematical relationships that connect the concentration of reactants to time, based on the reaction order. They are essentially equations derived from the differential rate laws, which describe how the rate depends on concentration.For a zero-order reaction, the integrated rate law is:\[[ ext{A}] = -kt + [ ext{A}]_0\]This equation indicates that the concentration decreases linearly with time, showing a constant rate.For a first-order reaction, the integrated rate law is:\[ ext{ln}[ ext{A}] = -kt + ext{ln}[ ext{A}]_0\]Here, the natural logarithm of concentration decreases linearly with time. This reflects an exponential decay of the concentration.For a second-order reaction, the integrated rate law is:\[rac{1}{[ ext{A}]} = kt + rac{1}{[ ext{A}]_0}\]In this scenario, the inverse of the concentration increases linearly with time.Using integrated rate laws allows us to predict how concentrations change over time and can be invaluable for planning chemical processes, optimizing reaction conditions, and ensuring safety in chemical manufacturing. Once the form of the rate law is determined and the rate constant is known, chemists can efficiently compute the expected concentrations of reactants at any given time.